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1) The absorbance of a 20 mM sample reads 1.5 using a 1 cmcell. What is the mola

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Question

1) The absorbance of a 20 mM sample reads 1.5 using a 1 cmcell. What is the molar absorptivity of the sample. 2a) If the sample concentration is reduced by 20%, what is thenew absorbance obtained? 2b) If the sample cell in the question (2a) is doubled insize, what should the new absorbance reading for the samplebe. 1) The absorbance of a 20 mM sample reads 1.5 using a 1 cmcell. What is the molar absorptivity of the sample. 2a) If the sample concentration is reduced by 20%, what is thenew absorbance obtained? 2b) If the sample cell in the question (2a) is doubled insize, what should the new absorbance reading for the samplebe.

Explanation / Answer

   a) the given data is Concentration C =20mM                               Absorbance     A= 1.5                        lengthof the sample   l =1cm the molar absorptivity of the sample = A /C l                                                            =  1.5 /20 mM *1cm   1mM=0.1 cm                                                            = 1.5 / 2                                                             =0.75 b)If the sample concentration is reduced by 20%                                                              =20mM - 20/100mM                                                              =(20-0.2)mM                                                               =19.8mM                            the new absorbance A = C l                                                         =0.75 *19.8mM* 1cM       1mM=0.1 cm                                                          = 0.75 *(19.8/10)                                                             = 1.485 c) If the sample cell in the question (2a) is doubled insize that means l= 2cm      the new absorbance reading for thesample be A = C l                                                                               = 0.75 *19.8mM* 2 cm                                                                                =0.75 *(19.8/10) *2                                                                                = 2.97